Chapter 12, Problem 78AE

(a)

Interpretation Introduction

Interpretation: The pressure of ${\text{PCl}}_{\text{5}}$ in undissociated form is to be calculated. The partial pressure, the total pressure and the percent dissociation of ${\text{PCl}}_{\text{5}}$ at equilibrium is to be calculated.

Concept introduction: The equilibrium constant $K_{\text{p}}$ describes the ratio of the reactant to the product on the equilibrium conditions in terms of the partial pressure..

At equilibrium sum of all the partial pressure is equal to the total pressure.

Law of mass action is applicable on the equilibrium reactions.

The molar mass of ${\text{PCl}}_{\text{5}}=208.23\text{g}\cdot {\text{mol}}^{\text{-1}}$.

Dissociation is also known as degree of dissociation which tells that dissociated amount of the species at a particular time.

To determine: The pressure of ${\text{PCl}}_{\text{5}}$ in undissociated form.

Answer to Problem 78AE

Answer

The pressure of ${\text{PCl}}_{\text{5}}$ in undissociated form is $\underset{\_}{1.16atm}$.

Explanation of Solution

Explanation

The pressure of ${\text{PCl}}_{\text{5}}$ in undissociated form is $\underset{\_}{1.16atm}$.

Given

The reactions are given as,

${\text{PCl}}_{\text{5}}{}_{\left(s\right)}\rightleftharpoons {\text{PCl}}_{\text{5}}{}_{\left(g\right)}+{\text{Cl}}_2{}_{\left(g\right)}$

The equilibrium constant $K_{\text{p}}=11.5$ at $600\text{K}$.

Mass of ${\text{PCl}}_{\text{5}}=2.450\text{g}$.

Volume of the ${\text{PCl}}_{\text{5}}$ $=500\text{ml}$.

Moles of the ${\text{PCl}}_{\text{5}}$ is determined by the formula,

$n=\frac{m}{M}$

Where,

• $n$ is the number of moles.
• $m$ is the given mass.
• $M$ is the molar mass.

Substitute the values in the above equation.

$\begin{array}{c}n=\frac{m}{M}\\ =\frac{2.450\text{g}}{208.23\text{g}\cdot {\text{mol}}^{\text{-1}}}\\ =0.01176\text{mol}\end{array}$

Volume of the ${\text{PCl}}_{\text{5}}$ gas in Liter $=0.50\text{L}$

The pressure of ${\text{PCl}}_{\text{5}}$ in undissociated form is given by the formula,

$P=\left(\frac{n}{V}\right)\text{R}T$

Where,

• $P$ is the pressure.
• $V$ is the volume in L.
• $\text{R}$ is the gas constant.
• $T$ is the temperature.

Substitute the values in the above equation.

$\begin{array}{c}P=\left(\frac{n}{V}\right)\text{R}T\\ =\left(\frac{0.01176\text{mol}}{0.50\text{L}}\right)0.08206\text{L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{\text{-1}}\times 600\text{K}\\ =\underset{\_}{1.16atm}\end{array}$

Conclusion

The pressure of ${\text{PCl}}_{\text{5}}$ in undissociated form is $\underset{\_}{1.16atm}$.

(b)

Interpretation Introduction

Interpretation: The pressure of ${\text{PCl}}_{\text{5}}$ in undissociated form is to be calculated. The partial pressure, the total pressure and the percent dissociation of ${\text{PCl}}_{\text{5}}$ at equilibrium is to be calculated.

Concept introduction: The equilibrium constant $K_{\text{p}}$ describes the ratio of the reactant to the product on the equilibrium conditions in terms of the partial pressure..

At equilibrium sum of all the partial pressure is equal to the total pressure.

Law of mass action is applicable on the equilibrium reactions.

The molar mass of ${\text{PCl}}_{\text{5}}=208.23\text{g}\cdot {\text{mol}}^{\text{-1}}$.

Dissociation is also known as degree of dissociation which tells that dissociated amount of the species at a particular time.

To determine: The partial pressure of ${\text{PCl}}_{\text{5}}$ at equilibrium.

Answer to Problem 78AE

Answer

The partial pressure of ${\text{PCl}}_{\text{5}}$ at equilibrium is $\underset{\_}{0.10atm}$.

Explanation of Solution

Explanation

The partial pressure of ${\text{PCl}}_{\text{5}}$ at equilibrium is $\underset{\_}{0.10atm}$.

The initial partial pressure of ${\text{PCl}}_{\text{5}}$ is calculated in step 1 $=1.16\text{atm}$ and the change in partial pressure is $x$.

The (ICE-chart) that is the initial pressure, change in pressure and equilibrium pressure is given as,

$\begin{array}{cccc}& {\text{PCl}}_{\text{5}}{}_{\left(\text{g}\right)}\rightleftharpoons & {\text{PCl}}_{\text{3}}{}_{\left(\text{g}\right)}& {\text{Cl}}_{\text{2}}{}_{\left(\text{g}\right)}\\ \text{Initial}\text{concentration}& \text{1}\text{.16}& \text{0}& \text{0}\\ \text{Change}\text{in}\text{concentration}& \text{-x}& \text{+x}& \text{+x}\\ \text{Equilibrium}\text{concentration}& \text{1}\text{.16-x}& \text{x}& \text{x}\end{array}$

The equilibrium constant in terms of partial pressure for the given reaction is given by the formula.

$k_{\text{p}}=\frac{P_{{\text{PCl}}_{\text{3}}}^{}{P}_{{\text{Cl}}_2}}{P_{{\text{PCl}}_5}^{}}$

Where,

• $P_{{\text{PCl}}_{\text{3}}}^{}$ is the partial pressure of ${\text{PCl}}_{\text{3}}$.
• $P_{{\text{Cl}}_2}^{}$ is the partial pressure of ${\text{Cl}}_2$.
• $P_{{\text{PCl}}_5}^{}$ is the partial pressure of ${\text{PCl}}_{\text{5}}$.

Substitute the values of partial pressures from the ICE-table in the above equation.

$\begin{array}{c}{k}_{\text{p}}=\frac{P_{{\text{PCl}}_{\text{3}}}^{}{P}_{{\text{Cl}}_2}}{P_{{\text{PCl}}_5}^{}}\\ 11.5=\frac{\left(x\right)\left(x\right)}{\left(1.16-x\right)}\\ x=1.06\end{array}$

From the ICE table the equilibrium partial pressure of ${\text{PCl}}_5{}_{\left(g\right)}$ is calculated as,

$\begin{array}{c}{P}_{{\text{PCl}}_{\text{5}}}=1.16-x\\ =1.16-1.06\\ =\underset{\_}{0.10atm}\end{array}$

Conclusion

The partial pressure of ${\text{PCl}}_{\text{5}}$ at equilibrium is $\underset{\_}{0.10atm}$.

(c)

Interpretation Introduction

Interpretation: The pressure of ${\text{PCl}}_{\text{5}}$ in undissociated form is to be calculated. The partial pressure, the total pressure and the percent dissociation of ${\text{PCl}}_{\text{5}}$ at equilibrium is to be calculated.

Concept introduction: The equilibrium constant $K_{\text{p}}$ describes the ratio of the reactant to the product on the equilibrium conditions in terms of the partial pressure..

At equilibrium sum of all the partial pressure is equal to the total pressure.

Law of mass action is applicable on the equilibrium reactions.

The molar mass of ${\text{PCl}}_{\text{5}}=208.23\text{g}\cdot {\text{mol}}^{\text{-1}}$.

Dissociation is also known as degree of dissociation which tells that dissociated amount of the species at a particular time.

To determine: The total pressure in the bulb at equilibrium.

Answer to Problem 78AE

Answer

The total pressure of ${\text{PCl}}_{\text{5}}$ at equilibrium is $\underset{\_}{2.22atm}$.

Explanation of Solution

Explanation

The total pressure of ${\text{PCl}}_{\text{5}}$ at equilibrium is $\underset{\_}{2.22atm}$.

At equilibrium the partial pressure is equal to the sum of the individual pressure of the all gases. Therefore,

$P_{\text{total}}={\displaystyle \sum P_{\text{partial}}}$

Where,

• $P_{\text{total}}$ is the total pressure at equilibrium.
• $P_{\text{partial}}$ is the partial pressure of the gases.
• ${\displaystyle \sum }$ represents the sum of all partial pressures.

Substitute the given value of total pressure and value of partial pressure of all gases.

$\begin{array}{c}{P}_{\text{total}}={\displaystyle \sum P_{\text{partial}}}\\ ={\text{P}}_{{\text{PCl}}_{\text{5}}}+{\text{P}}_{{\text{PCl}}_3}+{\text{P}}_{{\text{Cl}}_2}\\ =0.10+1.062+1.062\\ =\underset{\_}{2.22atm}\end{array}$

Conclusion

The total pressure of ${\text{PCl}}_{\text{5}}$ at equilibrium is $\underset{\_}{2.22atm}$.

(d)

Interpretation Introduction

Interpretation: The pressure of ${\text{PCl}}_{\text{5}}$ in undissociated form is to be calculated. The partial pressure, the total pressure and the percent dissociation of ${\text{PCl}}_{\text{5}}$ at equilibrium is to be calculated.

Concept introduction: The equilibrium constant $K_{\text{p}}$ describes the ratio of the reactant to the product on the equilibrium conditions in terms of the partial pressure..

At equilibrium sum of all the partial pressure is equal to the total pressure.

Law of mass action is applicable on the equilibrium reactions.

The molar mass of ${\text{PCl}}_{\text{5}}=208.23\text{g}\cdot {\text{mol}}^{\text{-1}}$.

Dissociation is also known as degree of dissociation which tells that dissociated amount of the species at a particular time.

To determine: The percent dissociation of ${\text{PCl}}_{\text{5}}$ at equilibrium.

Answer to Problem 78AE

Answer

The percent dissociation of ${\text{PCl}}_{\text{5}}$ at equilibrium is $\underset{\_}{91.4\%}$.

Explanation of Solution

Explanation

The percent dissociation of ${\text{PCl}}_{\text{5}}$ at equilibrium is $\underset{\_}{91.4\%}$.

The initial moles of ${\text{PCl}}_{\text{5}}$ calculated in step 1 $=0.0117\text{mol}$

At equilibrium the number of moles calculated by the formula,

$n=\frac{PV}{\text{R}T}$

Substitute the values in the above equation.

$\begin{array}{c}n=\frac{PV}{\text{R}T}\\ =\frac{\left(0.10\text{atm}\right)\left(0.50\text{L}\right)}{0.08206\text{L}\cdot \text{atm}\cdot {\text{K}}^{\text{-1}}\cdot {\text{mol}}^{\text{-1}}\times 600\text{K}}\\ =0.001\text{mol}\end{array}$

The number of reacted moles is given as,

$\begin{array}{l}=0.0117-0.001\\ =0.0107\text{mol}\end{array}$

Therefore, the percent degree of dissociation of ${\text{PCl}}_{\text{5}}$ at equilibrium is given by the formula,

$\%\alpha =\frac{n_{\text{reacted}}}{n_{\text{initial}}}\times 100$

Where,

• $\%\alpha$ is the percent degree of dissociation.
• $n_{\text{reacted}}$ is the number of reacted moles.
• $n_{\text{initial}}$ is the initial number of moles.

Substitute the values in the above equation.

$\begin{array}{c}\%\alpha =\frac{n_{\text{reacted}}}{n_{\text{initial}}}\times 100\\ =\frac{0.0107}{0.0117}\times 100\\ =\underset{\_}{91.4\%}\end{array}$

Conclusion

The percent dissociation of ${\text{PCl}}_{\text{5}}$ at equilibrium is $\underset{\_}{91.4\%}$.